Problem: Find $\lim_{x\to0}\dfrac{1-e^{x}}{\ln\left(2-e^{x}\right)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac2e$ (Choice B) B $1$ (Choice C) C $0$ (Choice D) D The limit doesn't exist.
Substituting $x=0$ into $\dfrac{1-e^{x}}{\ln\left(2-e^{x}\right)}$ results in the indeterminate form $\dfrac{0}{0}$. Furthermore, as the expression involves mixed function types, it's not possible to manipulate it algebraically in a way that will help us find the limits. Therefore, we should use L'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to0}\dfrac{1-e^{x}}{\ln\left(2-e^{x}\right)} \\\\ &=\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[1-e^{x}]}{\dfrac{d}{dx}[\ln\left(2-e^{x}\right)]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 0}\dfrac{-e^x}{\left(\dfrac{1}{2-e^x}\right)(-e^x)} \\\\ &=\dfrac{-e^{(0)}}{\left(\dfrac{1}{2-e^{(0)}}\right)(-e^{(0)})} \gray{\text{Substitution}} \\\\ &=1 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[1-e^{x}]}{\dfrac{d}{dx}[\ln\left(2-e^{x}\right)]}$ actually exists. In conclusion, $\lim_{x\to0}\dfrac{1-e^{x}}{\ln\left(2-e^{x}\right)}=1$.